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February 12th, 2019, 02:58 PM   #1
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Rare Even in Pinochle game

I am not an extremely experienced Pinochle player, but I would estimate that I have played thousands of hands. WhatvI am asking about seems extremely unlikely, and the answer that I got for this question seems questionable so....

I'm trying to determine the odds of a seemingly ultra rare event in a Pinochle game. I just don't know how to go about finding the answer.

For clarification in case you are unfamiliar with the game, the deck consists of 80 cards: four each A, 10, K, Q, J in each of the four suits for a total of 20 clubs, 20 diamonds, 20 hearts and 20 spades. A "run" is A, 10, K, Q, J in the same suit.

What are the odds, with 4 players that one player is dealt a double run (i.e.two runs of the same suit), and another player is dealt a run in the same suit in the same hand?

In all the Pinochle hands that I have played, I can't remember anyone even getting a double run, let alone someone else also getting the same run.

This is not for any academic work. This is just a unbelievable situation that happened in a card game with some friends.

I wouldn't mind an explanation of the procedure to determine the answer if you have the time.

Last edited by jk91977; February 12th, 2019 at 03:31 PM.
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February 12th, 2019, 04:26 PM   #2
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From my understanding of Pinochle and probability, which are both entry level, getting even one run would be a very rare occurrence. To my understanding, the probability of getting a single run would be 256/1,502,501

This is because your first card can be any card in the deck, as the cards the deck is made up of only consist of the A 10 K Q and J, all of which can be used to form a run. So the probability of you getting a correct first card is 80/80, or 1.

Your second card needs to be of the same suit, thus ruling out 60 cards. However, your next card cannot be the same one as the first card you picked, ruling out another 4, therefore, you have a 16/79 (about 20%) chance of getting the required second card.

For the third card, you only have 12 options, as it needs to be one of the 20 cards within the suit, but can't be a duplicate of the ones before it. This means you have used up 8 cards within the suit, leaving 12 cards that could be picked. You have a 12/78 (about 15%) chance of picking the required third card.

For the fourth card, you now only have 8 options, as it needs to be one of the as it needs to be one of the 20 cards within the suit, but can't be a duplicate of the ones before it. This means you have used up 12 cards within the suit, leaving only 8 cards that could be picked. You have a 8/77 (about 10%) chance of picking the required fourth card.

For the fifth and final card, you now only have 4 options, as you have used up all the other types of cards. you have a 4/76 (about 5%) chance of picking the required fifth card.

Now, you multiply the probability of getting each card by one another (if you want to know why, please look it up, as it is a long explanation) and you get the probability of getting a single run, being 256/1,502,501

Keep in mind that this probability representation would be if you picked your cards first, as I am not taking into account already used cards, which would make the situation even more unlikely.

I am not sure how you would figure out getting two runs in one hand or even having another person dealt a run, but I hope this information helps. Again, this is based on entry level knowledge, so my understanding might not be fully accurate.
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February 13th, 2019, 12:57 PM   #3
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This seems to be an unusual pinochle deck. The ones I played with had 9,10,J,Q,K,A twice in each suit for a deck of 48 cards
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February 14th, 2019, 04:32 AM   #4
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Quote:
Originally Posted by Diak42 View Post
From my understanding of Pinochle and probability, which are both entry level, getting even one run would be a very rare occurrence. To my understanding, the probability of getting a single run would be 256/1,502,501

This is because your first card can be any card in the deck, as the cards the deck is made up of only consist of the A 10 K Q and J, all of which can be used to form a run. So the probability of you getting a correct first card is 80/80, or 1.

Your second card needs to be of the same suit, thus ruling out 60 cards. However, your next card cannot be the same one as the first card you picked, ruling out another 4, therefore, you have a 16/79 (about 20%) chance of getting the required second card.

For the third card, you only have 12 options, as it needs to be one of the 20 cards within the suit, but can't be a duplicate of the ones before it. This means you have used up 8 cards within the suit, leaving 12 cards that could be picked. You have a 12/78 (about 15%) chance of picking the required third card.

For the fourth card, you now only have 8 options, as it needs to be one of the as it needs to be one of the 20 cards within the suit, but can't be a duplicate of the ones before it. This means you have used up 12 cards within the suit, leaving only 8 cards that could be picked. You have a 8/77 (about 10%) chance of picking the required fourth card.

For the fifth and final card, you now only have 4 options, as you have used up all the other types of cards. you have a 4/76 (about 5%) chance of picking the required fifth card.

Now, you multiply the probability of getting each card by one another (if you want to know why, please look it up, as it is a long explanation) and you get the probability of getting a single run, being 256/1,502,501

Keep in mind that this probability representation would be if you picked your cards first, as I am not taking into account already used cards, which would make the situation even more unlikely.

I am not sure how you would figure out getting two runs in one hand or even having another person dealt a run, but I hope this information helps. Again, this is based on entry level knowledge, so my understanding might not be fully accurate.
I failed to mention that there are 20 cards in your hand.
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